LeetCode - Next Permutation

題目:spa

Implement next permutation, which rearranges numbers into the lexicographically next greater permutation of numbers.code

If such arrangement is not possible, it must rearrange it as the lowest possible order (ie, sorted in ascending order).blog

The replacement must be in-place, do not allocate extra memory.element

Here are some examples. Inputs are in the left-hand column and its corresponding outputs are in the right-hand column.
1,2,31,3,2
3,2,11,2,3
1,1,51,5,1string

思路:it

摘自Stack Overflow:io

Find the highest index i such that s{i} < s{i+1}. If no such index exists, the permutation is the last permutation. Find the highest index j > i such that s{j} > s{i}. Such a j must exist, since i+1 is such an index. Swap s{i} with s{j}. Reverse all the order of all of the elements* after index i.ast

package string;

public class NextPermutation {

    public void nextPermutation(int[] nums) {
        int len;
        if (nums == null || (len = nums.length) < 2) return;
        int index1 = -1;
        for (int i = 0; i < len - 1; ++i) {
            if (nums[i + 1] > nums[i])
                index1 = i;
        }
        
        if (index1 != -1) {
            int index2 = index1 + 1;
            for (int j = index2 + 1; j < len; ++j) {
                if (nums[j] > nums[index1])
                    index2 = j;
            }
            
            swap(nums, index1, index2);
            reverse(nums, index1 + 1, len);
        } else {
            reverse(nums, 0, len);
        }
    }
    
    private void swap(int[] nums, int i, int j) {
        int tmp = nums[i];
        nums[i] = nums[j];
        nums[j] = tmp;
    }
    
    private void reverse(int[] nums, int start, int end) {
        for (int i = start; i < (end - start) / 2 + start; ++i) {
            swap(nums, i, end - i + start - 1);
        }
    }
    
    public static void main(String[] args) {
        // TODO Auto-generated method stub
        NextPermutation n = new NextPermutation();
        int[] nums = { 1,3,2 };
        n.nextPermutation(nums);
        for (int i = 0; i < nums.length; ++i)
            System.out.println(nums[i]);
    }

}
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